- #1
Meselwulf
- 125
- 0
I'm definitely messing up somewhere :P
The Larmor Equation is
\Delta H = \frac{2\mu}{\hbar Mc^2 e} \frac{\partial V(r_{ij})}{\partial r_{ij}} ( L \cdot S)
I derived an equation stated as
\nabla \times \vec{F}_{ij} \cdot \hat{n} = \nabla \times \frac{\partial V(r_{ij})}{\partial r_{ij}}
How is put in a simpler way from the original derivation, is imagine we have what we began with
\vec{F}_{ij} = \frac{\partial V(r_{ij})}{\partial r_{ij}} \hat{n}
Take the dot product of the unit vector on both sides gives
\vec{F}_{ij} \cdot \hat{n} = \frac{\partial V(r_{ij})}{\partial r_{ij}}
Now taking the curl of F is
\nabla \times \vec{F}_{ij} \cdot \hat{n} = \nabla \times \frac{\partial V(r_{ij})}{\partial r_{ij}}
In light of this equation however
\vec{F}_{ij} \cdot \hat{n} = \frac{\partial V(r_{ij})}{\partial r_{ij}}
I could rewrite the Larmor energy as
\Delta H = \frac{2\mu}{\hbar Mc^2 e}(\vec{F}_{ij} \cdot \hat{n}) L \cdot S
But I am sure many agree that is not very interesting.
(and this is really starting to make my brain cells burst), taking the curl of a force gives a F/length, however, the unit vector in this equation
\vec{F}_{ij} \cdot \hat{n} = \frac{\partial V(r_{ij})}{\partial r_{ij}}
cancels these out and what I have again is the force again... yes? ... very circular... since this would be true, then we know what the force is anyway given earlier:
\vec{F}_{ij} = \frac{\partial V(r_{ij})}{\partial r_{ij}} \hat{n}
Now, just a moment ago I found the Larmor energy written as
\Delta H = \frac{2\mu}{\hbar Mc^2 e} \frac{1}{r}\frac{\partial V(r_{ij})}{\partial r_{ij}} ( L \cdot S)
Notice the 1/r term which is not in my original case. If that where true, and we plug in my force example again
\Delta H = \frac{2\mu}{\hbar Mc^2 e} \frac{1}{r}\frac{\partial V(r_{ij})}{\partial r_{ij}}\hat{n} ( L \cdot S)
(If I am doing this right) the unit vector would cancel out with the radius term and what would be left with is
\Delta H = \frac{2\mu}{\hbar Mc^2 e} \frac{\partial V(r_{ij})}{\partial r_{ij}} ( L \cdot S)
Now if that wasn't confusing I don't know what is.
I've either had it right from the beginning, or I have made a tiny mistake which is making a huge impact on my understanding of my own equation. Any help would be gladly appreciated.
The Larmor Equation is
\Delta H = \frac{2\mu}{\hbar Mc^2 e} \frac{\partial V(r_{ij})}{\partial r_{ij}} ( L \cdot S)
I derived an equation stated as
\nabla \times \vec{F}_{ij} \cdot \hat{n} = \nabla \times \frac{\partial V(r_{ij})}{\partial r_{ij}}
How is put in a simpler way from the original derivation, is imagine we have what we began with
\vec{F}_{ij} = \frac{\partial V(r_{ij})}{\partial r_{ij}} \hat{n}
Take the dot product of the unit vector on both sides gives
\vec{F}_{ij} \cdot \hat{n} = \frac{\partial V(r_{ij})}{\partial r_{ij}}
Now taking the curl of F is
\nabla \times \vec{F}_{ij} \cdot \hat{n} = \nabla \times \frac{\partial V(r_{ij})}{\partial r_{ij}}
In light of this equation however
\vec{F}_{ij} \cdot \hat{n} = \frac{\partial V(r_{ij})}{\partial r_{ij}}
I could rewrite the Larmor energy as
\Delta H = \frac{2\mu}{\hbar Mc^2 e}(\vec{F}_{ij} \cdot \hat{n}) L \cdot S
But I am sure many agree that is not very interesting.
(and this is really starting to make my brain cells burst), taking the curl of a force gives a F/length, however, the unit vector in this equation
\vec{F}_{ij} \cdot \hat{n} = \frac{\partial V(r_{ij})}{\partial r_{ij}}
cancels these out and what I have again is the force again... yes? ... very circular... since this would be true, then we know what the force is anyway given earlier:
\vec{F}_{ij} = \frac{\partial V(r_{ij})}{\partial r_{ij}} \hat{n}
Now, just a moment ago I found the Larmor energy written as
\Delta H = \frac{2\mu}{\hbar Mc^2 e} \frac{1}{r}\frac{\partial V(r_{ij})}{\partial r_{ij}} ( L \cdot S)
Notice the 1/r term which is not in my original case. If that where true, and we plug in my force example again
\Delta H = \frac{2\mu}{\hbar Mc^2 e} \frac{1}{r}\frac{\partial V(r_{ij})}{\partial r_{ij}}\hat{n} ( L \cdot S)
(If I am doing this right) the unit vector would cancel out with the radius term and what would be left with is
\Delta H = \frac{2\mu}{\hbar Mc^2 e} \frac{\partial V(r_{ij})}{\partial r_{ij}} ( L \cdot S)
Now if that wasn't confusing I don't know what is.
I've either had it right from the beginning, or I have made a tiny mistake which is making a huge impact on my understanding of my own equation. Any help would be gladly appreciated.